Question

If each term of a geometric progression $a_1, a_2, a_3, \dots$ with $a_1 = \frac{1}{8}$ and $a_2 \neq a_1$, is the arithmetic mean of the next two terms and $S_n = a_1 + a_2 + \dots + a_n$, then $S_{20} - S_{18}$ is equal to

• A. $2^{15}$
• B. $-2^{18}$
• C. $2^{18}$
• D. $-2^{15}$

Answer 1

Answer: (D)

$-2^{15}$

Answer 2

Let the r-th term of the geometric progression (GP) be $ar^{r-1}$.

Step 1. Given condition: Since each term is the arithmetic mean of the next two terms, we have:
$2a_r = a_{r+1} + a_{r+2}$
Substituting the terms, this becomes:
$2ar^{r-1} = ar^r + ar^{r+1}$
Dividing by $ar^{r-1}$ (assuming $a \neq 0$), we get:
$2 = r + r^2$

Step 2. Solve for $r$: Rearranging, we have:
[$r^2 + r - 2 = 0$
Factoring, we get:
$(r - 1)(r + 2) = 0$
Thus, $r = 1$ or $r = -2$. Since $a_2 \neq a_1$, $r \neq 1$, so $r = -2$.

Step 3. Calculate $S_{20} - S_{18}$: Since the sum of the first $n$ terms of a GP is given by
$S_n = \frac{a(r^n - 1)}{r - 1},$
we find $S_{20}$ and $S_{18}$:
$S_{20} = \frac{1}{-2}((-2)^{20} - 1), \quad S_{18} = \frac{1}{-2}((-2)^{18} - 1).$

Therefore,
$S_{20} - S_{18} = \frac{1}{-2}((-2)^{20} - (-2)^{18})$
Simplifying, we get:
$S_{20} - S_{18} = -2^{15}.$
The Correct Answer is:$-2^{15}$.